Re: Re: Curious about method handling

> > String f() =3D "Hello World";
> >=20
> > class A {
> >   String greeting =3D "Hello World";
> >   String g() =3D this.greeting;
> > }
> >=20
> > void main(String[] args) {
> >   println( f );
> >   println( ( new A() ).g );
> > }

1) As you've seen, instead of applying f and printing the result, this
just prints a representation of the function f
   println( f );

Do this to apply the function
   println( f() );

2) These both mean the same
   println( new A().g );
   println( new A().g() );

So the interesting question is how do a get a reference to a non-static
method of class A.


This gives a compiler error
> repeatOnA( new A(), A.g );
Class test.A has no static method or static field named g


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